`I^(-)` reduces `IO_(3)^(-)` to `I_(2)` and itself oxidised to `I_(2)` in acidic medium. Thus, final reaction is

Explanation

To solve the problem, we need to write the balanced chemical reaction for the reduction of \( IO_3^- \) by \( I^- \) to form \( I_2 \) in acidic medium. Here’s a step-by-step breakdown of the solution:

### Step 1: Identify the reactants and products
The reactants are \( I^- \) (iodide ion) and \( IO_3^- \) (iodate ion). The products are \( I_2 \) (iodine) and water \( H_2O \) as we are in acidic medium.

### Step 2: Write the unbalanced equation
The unbalanced equation based on the information given is:
\[ I^- + IO_3^- \rightarrow I_2 \]

### Step 3: Determine oxidation states
- In \( IO_3^- \), iodine has an oxidation state of +5.
- In \( I_2 \) and \( I^- \), iodine has an oxidation state of 0 and -1, respectively.

### Step 4: Balance the iodine atoms
To balance the iodine atoms, we note that \( IO_3^- \) contains 1 iodine atom and \( I_2 \) contains 2 iodine atoms. Therefore, we need to balance the iodine:
\[ 5 I^- + IO_3^- \rightarrow 3 I_2 \]

### Step 5: Balance the oxygen atoms
The reactant \( IO_3^- \) has 3 oxygen atoms. To balance the oxygen atoms, we can add water molecules to the products:
\[ 5 I^- + IO_3^- + 6 H^+ \rightarrow 3 I_2 + 3 H_2O \]

### Step 6: Write the final balanced equation
The final balanced equation is:
\[ 5 I^- + IO_3^- + 6 H^+ \rightarrow 3 I_2 + 3 H_2O \]

### Final Reaction
Thus, the final reaction is:
\[ 5 I^- + IO_3^- + 6 H^+ \rightarrow 3 I_2 + 3 H_2O \]

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